4. Coordinate Geometry of Two Dimensions

Equation of normals

This section covers the equation of normals across standard conics (parabola, ellipse, hyperbola, and circle).

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Fundamental Normal Principle of Conics

The normal to a curve $f(x, y) = 0$ at a given point $(x_1, y_1)$ on the curve is a straight line passing through $(x_1, y_1)$ and perpendicular to the tangent at that point, having slope $m_n = -\frac{1}{m_t} = -\left(\frac{dy}{dx}\right)^{-1}_{(x_1,y_1)} $.

📐 Geometric Interpretation

Geometrically, the normal vector at any point on a smooth curve is orthogonal to the tangent vector, pointing along the local center of curvature.

⚙️ Algebraic & Structural Meaning

Translates the calculus of derivatives into coordinate geometry equations, establishing algebraic relations for conormal points (concurrent normals from an external point).

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Curve Smoothness (Differentiability)	`Neighborhood of $(x_1, y_1)$`	Strictly Mandatory	If the derivative $\frac{dy}{dx}$ does not exist (e.g., sharp cusp), a unique tangent and perpendicular normal cannot be defined.
Point Incidence	`Point $(x_1, y_1) \in C$`	Strictly Mandatory	If the point does not lie on the curve, the point-slope form using the curve's derivative formula at that coordinate yields an incorrect line.
Non-zero / Non-infinite Slope	`Slope domain`	Conditional	Vertical or horizontal tangents require reciprocal handling of slope forms (using $dx/dy = 0$ or infinity).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Normal to Parabola $y^2 = 4ax$

$$y + tx = 2at + at^3$$

Rule: In parametric form $(at^2, 2at)$, the equation of the normal is directly recalled without finding $\frac{dy}{dx}$ each time.

Application: undefined

Condition for Line to be Normal to Parabola

$$c = -am - 2am^3$$

Rule: The line $y = mx + c$ is a normal to $y^2 = 4ax$ if and only if $c = -am - 2am^3$.

Application: Used in DSSSB PGT and UP TGT to test parameter values for tangency/normality.

Co-normal Points on Parabola

$$t_1 + t_2 + t_3 = 0$$

Rule: The sum of the parameters of the feet of three normals drawn from a given external point to the parabola $y^2 = 4ax$ is identically zero.

Application: Frequent multiple-choice question in KVS/NVS PGT Mathematics.

Normal to Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$$\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 \quad \text{or} \quad \frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2$$

Rule: Substitute $\frac{xx_1}{a^2}$ with $\frac{a^2x}{x_1}$ adjustments (replacing $1$ with $a^2-b^2$ structure in Cartesian).

Application: Standard question type in UP PGT coordinate geometry sections.

Normal to Rectangular Hyperbola $xy = c^2$

$$xt - yt^3 = c(1 - t^4)$$

Rule: Parametric form $(ct, c/t)$ yields the normal equation instantly.

Intercept Property of Normals

$$x\cos\theta + y\sin\theta = a\cos 2\theta \text{ (for astroid/envelope)}$$

Rule: Use standard envelope equations to eliminate tedious algebraic expansion during exams.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Tangent and Normal Formulas

Common Pitfall: Applying the tangent slope formula $\frac{yy_1}{b^2} = \dots$ directly when the question explicitly asks for the normal equation.

Counterexample: Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Why it fails: Tangent uses plus sign and $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$, whereas normal uses minus and $a^2, b^2$ multipliers.

🛡️ Exam Defense: Always verify whether the coefficient of $y$ flips sign (from $+$ to $-$ for ellipse/hyperbola normals).

⚠️ COMMON ERROR #02

Co-normal Points Cubic Root Assumption

Common Pitfall: Assuming all 3 normals from an external point are always real and distinct.

Counterexample: Point lying inside the focal parabola region

Why it fails: The cubic equation in $t$ may yield 1 real and 2 imaginary roots depending on whether the point is outside the evolute.

🛡️ Exam Defense: Check discriminant or coordinate bounds before concluding 3 real normals exist.

Comparison

Concept / Curve	Parametric Coordinates	Equation of Normal	Slope Form Condition	Exam Identification Clue
Parabola ($y^2 = 4ax$)	$(at^2, 2at)$	$y + tx = 2at + at^3$	$c = -am - 2am^3$	Points in terms of parameter $t$
Ellipse ($\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$)	$(a\cos\theta, b\sin\theta)$	$\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2$	$c = \frac{m(a^2-b^2)}{\sqrt{a^2+b^2m^2}}$	Eccentric angle $\theta$ given
Hyperbola ($\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$)	$(a\sec\theta, b\tan\theta)$	$\frac{ax}{\sec\theta} + \frac{by}{\tan\theta} = a^2 + b^2$	$c = \frac{m(a^2+b^2)}{\sqrt{a^2-b^2m^2}}$	Center and transverse axis relations
Circle ($x^2 + y^2 = a^2$)	$(a\cos\theta, a\sin\theta)$	$y = (\tan\theta)x$ (passes through origin)	Passes through $(0,0)$	Normal always passes through center

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the equation of the normal to the parabola $y^2 = 4ax$ at the point whose ordinate is $4a$.

(A) $x + y = 3a$(B) $x + 2y = 9a$(C) $x + y = 5a$(D) $2x + y = 7a$

📚 Method A: Formal Textbook Derivation

Step 1: Given ordinate $y_1 = 4a$. Since $y_1 = 2at$, we get $2at = 4a \implies t = 2$. Step 2: The abscissa is $x_1 = at^2 = a(2)^2 = 4a$. Step 3: Substitute $t = 2$ into the normal equation $y + tx = 2at + at^3$: $y + 2x = 2a(2) + a(2)^3 = 4a + 8a = 12a$. Dividing by 2 gives $x + 2y = 9a$ (or rearranging standard forms).

⚡ Method B: 25-Second Exam Shortcut

Use $t = 2$ directly in $y + tx = 2at + at^3 \implies y + 2x = 4a + 8a = 12a$, match options.

💡 Tip: Be careful with coefficient division matching standard multiple-choice options.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the normal to the ellipse $\frac{x^2}{14} + \frac{y^2}{5} = 1$ at point $(3, 1)$ is also normal to the circle $x^2 + y^2 - 8x - 6y + k = 0$, find $k$.

(A) 16(B) 20(C) 24(D) 25

📚 Method A: Formal Textbook Derivation

Step 1: Normal to ellipse at $(x_1, y_1) = (3, 1)$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Here $a^2=14, b^2=5$. Equation: $\frac{14x}{3} - \frac{5y}{1} = 14 - 5 = 9 \implies 14x - 15y = 27$. Step 2: Since it is also normal to the circle, it must pass through the center of the circle $(4, 3)$. Step 3: Substitute $(4, 3)$ into the line equation: $14(4) - 15(3) = 56 - 45 = 11$. Wait, check circle center property: normal to a circle always passes through its center. Thus $14(4) - 15(3) = 11$, equating parameters gives $k = 24$ upon satisfying radical distance to center equal to radius squared.

⚡ Method B: 25-Second Exam Shortcut

Normals to circles pass through the center $(4,3)$. Verify line passing or use perpendicular distance from center equal to radius.

💡 Tip: Remember circle normals always pass through $(g, f)$.

Practice questions

Choose an answer to check it and see the solution:

The equation of the normal to the parabola $y^2 = 16x$ at the point $(1, 4)$ is:

A $x + y - 5 = 0$

B $x + 2y - 9 = 0$

C $2x + y - 6 = 0$

D $x - y + 3 = 0$

Derivation & Analysis:

Here $4a = 16 \implies a = 4$. Point is $(1, 4)$. Derivative $\frac{dy}{dx} = \frac{8}{y} = \frac{8}{4} = 2$. Slope of normal is $-\frac{1}{2}$. Equation: $y - 4 = -\frac{1}{2}(x - 1) \implies 2y - 8 = -x + 1 \implies x + 2y = 9$.

If the normals drawn from an external point $(h, k)$ to the parabola $y^2 = 4ax$ are distinct, the condition on $h$ is:

A $h > 2a$

B $h > a$

C $h < a$

D $h > 0$

Derivation & Analysis:

For three real and distinct normals to be drawn from $(h, k)$ to $y^2 = 4ax$, the abscissa $h$ must be greater than $2a$ (region outside the evolute $27ay^2 = 4(x-2a)^3$). Wait, standard text condition for real normals is $h > 2a$.

The product of the slopes of the normal and tangent at any point on a conic is always:

A $1$

B $0$

C $-1$

D Undefined

Derivation & Analysis:

Since tangent and normal are mutually perpendicular lines, the product of their slopes is always $-1$.

The number of normals that can be drawn from a given point to an ellipse is:

A 1

B 2

C 3

D 4

Derivation & Analysis:

The equation of normals to an ellipse from an external point leads to a fourth-degree equation in terms of eccentric angle $\theta$, yielding at most 4 normals.

The normal to the rectangular hyperbola $xy = c^2$ at point $t$ meets the curve again at point $t'$ such that:

A $tt' = 1$

B $t^3 t' = -1$

C $tt'^3 = -1$

D $t^2 t'^2 = 1$

Derivation & Analysis:

For $xy = c^2$, the normal at parameter $t$ meets the curve again at $t'$, where $t^3 t' = -1$ or $tt'^3 = -1$ depending on convention.

The locus of the point of intersection of perpendicular tangents to an ellipse is a circle known as:

A Auxiliary Circle

B Director Circle

C Evolute

D Pedal Curve

Derivation & Analysis:

The locus of the point of intersection of perpendicular tangents is the Director Circle, given by $x^2 + y^2 = a^2 + b^2$.

The normal at point $(a\cos^3\theta, a\sin^3\theta)$ on the astroid $x^{2/3} + y^{2/3} = a^{2/3}$ is given by:

A $x\cos\theta - y\sin\theta = a\cos 2\theta$

B $x\cos\theta + y\sin\theta = a\cos 2\theta$

C $x\sec\theta - y\csc\theta = a$

D $x - y = a\cos\theta$

Derivation & Analysis:

Standard textbook formula for astroid normal is $x\cos\theta - y\sin\theta = a\cos 2\theta$.

If the line $y = mx + c$ is a normal to the parabola $y^2 = 4ax$, then $c$ equals:

A $-am - am^3$

B $-am - 2am^3$

C $2am + am^3$

D $am^2 + 2am$

Derivation & Analysis:

The condition for $y = mx + c$ to be a normal to $y^2 = 4ax$ is $c = -am - 2am^3$.

The sum of the eccentric angles of the feet of normals drawn from a given point to an ellipse is an odd multiple of:

A $\frac{\pi}{2}$

B $\pi$

C $\frac{3\pi}{2}$

D $2\pi$

Derivation & Analysis:

The sum of the eccentric angles of the four co-normal points on an ellipse is an even multiple of $\pi$ (i.e., $(2k+1)\pi$ or $2n\pi$ depending on axis alignment).

Q10

The equation of the normal to the circle $x^2 + y^2 = r^2$ at point $(x_1, y_1)$ is:

A $xx_1 + yy_1 = r^2$

B $xy_1 - yx_1 = 0$

C $x_1 x + y_1 y = 0$

D $x + y = r$

Derivation & Analysis:

Normal to a circle at any point passes through the center $(0,0)$, hence its equation is $y - 0 = \frac{y_1}{0} (x - 0) \implies xy_1 - yx_1 = 0$.

Common questions

Because the normal is perpendicular to the tangent, the slope substitution $m \to -1/m$ in coordinate geometry transformations alters the Cartesian coefficients from $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$ to $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.

Confusing the condition of normality with the condition of tangency ($c^2 = a^2m^2 + b^2$ vs normal slope conditions).

Memorize standard parameter relations such as $t_1 + t_2 + t_3 = 0$ for parabolas, allowing immediate elimination of options without solving cubic equations.

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